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Statement

Setting as on Theorem 2.1: equation (1) is $n/2^n=\sum_{i=1}^{k}a_i/2^{a_i}$ with $k>1$ and $a_1<\cdots<a_k$, and $N(k)$ is the number of its solutions with $k$ terms (p. 6).

Proposition 3.1 (p. 6). If

$$k\equiv 10131316054712759135960334995313053617046 \pmod{20263657997642451746458664712008831939580},$$

then (1) has at least five solutions, that is $N(k)\ge5$.

The paper conjectures (p. 8, Conjecture 3.3) that $\limsup_{k\to+\infty}N(k)=\infty$, and (Conjecture 3.2) that the set of positive $u$ for which the congruence (2) below is solvable is infinite.

The five solutions (pp. 6--8). For an integer $u\ge0$, the choice $a_i=n+i$ ($i\le k-2$), $a_{k-1}=n+k+u$, $a_k=n+k+u+1$ solves (1) exactly when

$$n=2^{k-1}-k+\frac{3\cdot2^{k-1}+3u+1}{2^{u+3}-3}$$

is an integer, that is when $3\cdot2^{k-1}+3u+1\equiv0\pmod{2^{u+3}-3}$ (congruence (2)); the solvable $k$ for a given $u$ form residue classes modulo the order $r$ of $2$ modulo $2^{u+3}-3$. Table 1 (p. 7) lists one solution $k_0$ and $r$ for each of the 16 values $u\le120$ for which (2) is solvable. Four values of $u$ give four solutions for one $k$, and the fifth is $n=2^{k+1}-k-2$ with $a_i=n+i$ for all $i$.

Checked here in exact arithmetic: the printed modulus is the least common multiple of the orders $r=28,\ 4092,\ 1116130$ and $2535300206192230667655098198606$ of the rows $u=2,9,22,99$ of Table 1, every $k$ in the printed residue class satisfies (2) for these four values, and its least positive element satisfies (2) for no other $u\le120$. The proof's text (p. 8) names $u=2,9,55,99$ instead, and its displayed system and the values $x_1,\ldots,x_4$ lead to a different common value of $k$, which satisfies (2) for $u=2,9,55,99$ and not for $u=22$. Either way four values of $u$ apply, so the proposition as printed holds with $u=2,9,22,99$.

Source. Sz. Tengely, M. Ulas and J. Zygadło, On a Diophantine equation of Erdős and Graham, J. Number Theory 217 (2020), 445--459, doi:10.1016/j.jnt.2020.05.006, read in arXiv:2008.01501v1 as identified on the source card; labels and pages are that preprint's. Proposition 3.1 on p. 6, the derivation of (2) on pp. 6--7, Table 1 and the proof on pp. 7--8, Conjectures 3.2 and 3.3 on p. 8.

Read depth. Claims checked: the statement, Table 1's rows for $u=2,9,22,55,99$ and the proof's constants were read on the page images and recomputed as described above; the derivation of the formula for $n$ was checked on the case $u=0$, $k=4$, which gives the solution $(9;\,10,11,13,14)$ of Theorem 2.5. The paper's search over subsets of Table 1 was not rerun. Nothing here is independently reviewed.

Proof pointer

Pages 6--8. Rewriting (2) as $2^{k-1}\equiv(-3u-1)/3\pmod{2^{u+3}-3}$ turns it into a discrete logarithm problem for each $u$. Writing $f_i(x)=r_ix+k_i$ for the rows of Table 1, a $k$ shared by $m$ of the progressions gives $m$ solutions of the four-term-tail shape; the paper reports that no five rows have a common value and that exactly six sets of four rows do, and solves one such linear system by the Chinese remainder theorem.

Dependencies

Theorem 2.1 (Remark 2.2) for the fifth solution.

Bears on

• Problem 261: the proposition counts representations of $n/2^n$ with a fixed number of terms, over varying $n$. The problem asks which $n$ admit some representation and about rationals with $2^{\aleph_0}$ infinite representations; the proposition answers neither.